## Algebraic factor identities

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• Message 3 From: Andrey Kulsha Andrey_601@tut.by Date: Tue May 16, 2006 0:08pm(PDT) Subject: Re: A surprising algebraic factorization ...
Message 1 of 2 , May 17 8:37 AM
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Message 3
From: "Andrey Kulsha" Andrey_601@...
Date: Tue May 16, 2006 0:08pm(PDT)
Subject: Re: A surprising algebraic factorization

> 4 x^4 + y^4 = ( 2 x^2 - 2 x y + y^2) ( 2 x^2 + 2 x y + y^2)

http://xyyxf.at.tut.by/aurifeuillean.pdf

Which contains many identities such as

A^2 + B^2 = {1, 1}^2  2AB

Thanks.

I figure out the notation thusly.

(A + B)^2 = {1, 1}^2

The second identity in the http://xyyxf.at.tut.by/aurifeuillean.pdf

Is

(A^3 + B^3)/((A+B) = {1, 1}^2 - 3 A B

A^3 + B^3 = [ {1, 1}^2 - 3 A B ] (A + B) = {1,1}^2 - 3 A B (A + B)

(A + B)^3 = {1,1}^2 (A + B)

(A + B)^2 = {1,1}^2

All of the identities in that page are based on the difference of odd powers

The identity

> 4 x^4 + y^4 = ( 2 x^2 - 2 x y + y^2) ( 2 x^2 + 2 x y + y^2)

is surprising in that the difference of squares is the same as the sum of
4th powers.
• ... 2x^2 = a y^2 = b So 2ab is a perfect square, and we have an Aurifeuillean factorization you noticed. If 3ab, 5ab, 6ab, 7ab, 10ab, etc. is a perfect square,
Message 2 of 2 , May 17 12:33 PM
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> The identity
>
> > 4 x^4 + y^4 = ( 2 x^2 - 2 x y + y^2) ( 2 x^2 + 2 x y + y^2)
>
> is surprising in that the difference of squares is the same as the sum of
> 4th powers.

2x^2 = a
y^2 = b
So 2ab is a perfect square, and we have an Aurifeuillean factorization you noticed.

If 3ab, 5ab, 6ab, 7ab, 10ab, etc. is a perfect square, we'll be able to use the corresponding line in http://xyyxf.at.tut.by/aurifeuillean.pdf to obtain Aurifeuillean factors.

Here {1, 5, 7, 5, 1} means (a^4 + 5*a^3*b + 7*a^2*b^2 + 5*a*b^3 + b^4) and so on.

WBR,
Andrey

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